EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 7, No. 4, 2014, 429-436
ISSN 1307-5543 – www.ejpam.com

Regularity of the Rees and Associated Graded Modules

Naser Zamani

Faculty of Mathematical Sciences, University of Mohaghegh Ardabili, Ardabil, Iran

Abstract. Let A be a Noetherian ring and b be an ideal of A. Let E be a finitely generated A-module.
It is shown that there is a close relationship between the cohomological invariants of the associated
graded module of E with respect to b and the Rees module of E associated to b. Also a formula for the
regularity of the Rees module of E associated to b will be given.

2010 Mathematics Subject Classifications: 13A99, 13D45

Key Words and Phrases: associated graded rings and modules, graded local cohomology, reduction
number, filter regular sequence

1. Introduction

Let S = ⊕n≥0Sn be a finitely generated standard graded algebra over a Noetherian commu-
tative ring S0. We denote by S+ = ⊕n≥1Sn the ideal generated by the homogeneous elements
of positive degree of S. For a graded S-module L, the homogeneous part of degree n of L,
is denoted by Ln, and L(t) is the same module L shifted by t. The end of L is defined by
end(L) = max{n : Ln 6= 0}, and end(0) = −∞ by convention. For each i ≥ 0, the ith local
cohomology module H i

S+
(L) of a graded S-module L supported in S+ is also a graded S-module

in a natural way and H i
S+
(L)n is a finitely generated S0-module for all i ≥ 0 and all n, and it is

zero for for large values of n (see [1, Chapter 15]). Following [3], we put ai(L) = end(H i
S+
(L)).

Then the regularity of L is defined by reg(L) =max{ai(L) + i : i ≥ 0}.

Let A be a Noetherian commutative ring and b an ideal of A. Let E be a finitely gener-
ated A-module. We denote by Rb(E) = ⊕n≥0b

nE the Rees module of E associated to b and
by Gb(E) = ⊕n≥0b

nE/bn+1E = Rb(E)/bRb(E) the associated graded module of E with respect
to b. In the case E = A, these modules are denoted by R(b) and G(b) = R(b)/bR(b) respectively.

Recall from [2, Definition 4.6.4] that an ideal a ⊆ b is called a reduction of b with respect
to E if Rb(E) is a finitely generated R(a)-module, or equivalently, if br+1E = abr E for some

Email address: naserzaka@yahoo.com

http://www.ejpam.com 429 c© 2014 EJPAM All rights reserved.



N. Zamani / Eur. J. Pure Appl. Math, 7 (2014), 429-436 430

r ≥ 0. The least such r is denoted by ra(b, E).

This paper is divided into 3 sections. In section 2 we prepare some results related to the
Castelnuovo regularity of a graded module, from which we prove in theorem 1 that reg(L) can
be characterized in terms of a minimal reduction of S+ with respect to L, which is generated by
an S+-filter regular sequence of homogeneous elements of degree 1, for L. In section 3, using
the ideas of [5], we will show that there is a close relationship between the invariants ai(Rb(E))
and ai(Gb(E)), from which we can easily derive the formula reg(Rb(E)) = reg(Gb(E)). Also
we give a formula for the number reg(Rb(E)) in Corollary 4.

2. Preliminaries

From now on assume that L is finitely generated. Let f = f1, . . . , fh be a sequence of
homogeneous elements of S. We call f1, . . . , fh an S+-filter regular sequence for L if for all
i = 1, . . . , h

fi /∈
⋃

p∈AssS(L/( f1,..., fi−1)L)\V (S+)
p,

where V (S+) is the set of all prime ideals of S containing S+ and for an S-module X , AssS(X )
denotes the set of all associated prime ideals of X . We define

e(f, L) = sup{end((( f1, . . . , fi−1)L :L fi)/( f1, . . . , fi−1)L) : i = 1, . . . , h}.

Then by [1, 18.3.8], f1, . . . , fh is an S+-filter regular sequence on L if and only if e(f, L)<∞.

It will be crucial to understand how the invariants ai(L) behave with respect to S+-filter
regular sequences for L. This relationship was illuminated by Trung in the following lemma.
Because of its importance in our argument, we supply the proof along the statement.

Lemma 1 ([6, Lemma 2.3]). Let f ∈ S1 be a homogeneous S+-filter regular element for L. Then
for all i ≥ 0,

ai+1(L) + 1≤ ai(L/ f L)≤max{ai(L), ai+1(L) + 1}.

Proof. Note that by the statement after the definition of an S+-filter regular sequence for L,
H0

S+
(0 :L f ) = (0 :L f ) and hence H i

S+
(0 :L f ) = 0 for all i ≥ 1. Then from the exact sequence

0 −→ (0 :L f ) −→ L −→ L/(0 :L f ) −→ 0,

we see that H i
S+
(L)∼= H i

S+
(L/(0 :L f )) for all i ≥ 1. Now, from the exact sequence

0 −→ L/(0 :L f )
f
−→ L(1) −→ L(1) −→ 0,

we obtain the exact sequence

H i
S+
(L)n+1→ H i

S+
(L/ f L)n+1→ H i+1

S+
(L)n→ H i+1

S+
(L)n+1,

for each i ≥ 0 and n ∈ Z. Analyzing these sequences easily yields the desired inequalities.



N. Zamani / Eur. J. Pure Appl. Math, 7 (2014), 429-436 431

Lemma 2. Let f= f1, . . . , fh be an S+-filter regular sequence of homogeneous elements of degree
1 for L. Then:

(i) e(f, L) =max{ai(L) + i : i = 0, . . . , h− 1},

(ii) for all 0≤ t ≤ h,

max{ai(L)+i : i = 0, . . . , t}=max{end((( f1, . . . , ft)L :L S+)/( f1, . . . , ft)L) : i = 0, . . . , t}.

Proof. (i) We prove by induction on h≥ 1. Since (0 :L f1) ⊆ ∪n≥1(0 :L S+
n) and

f1H0
S+
(L)a0(L) ⊆ H0

S+
(L)a0(L)+1 = 0,

thus e( f1, L) = a0(L) and the case h = 1 is immediate. So let h > 1. Let L̄ = L/ f1 L and
f̄= f̄2, . . . , f̄h in S̄ = S/( f1). By induction and using Lemma 1, we have

max{ai(L) + i : i = 1, . . . , h− 1} ≤e(̄f, L̄) =max{ai(L/ f1 L) + i : i = 0, . . . , h− 2}
≤max{ai(L) + i : i = 0, . . . , h− 1}.

Now since e(f, L) =max{e( f1, L), e(̄f, L̄)}, the result follows.

(ii) Using Lemma 1 repeatedly, we deduce that

ai(L) + i ≤ a0(L/( f1, . . . , fi)L)≤max{a j(L) + j : j = 0, . . . , i}.

From this it follows that for t ≤ h,

max{ai(L) + i : i = 0, . . . , t}=max{a0(L/( f1, . . . , fi)L) : i = 0, . . . , t}.

Set a = a0(L/( f1, . . . , fi)L). We have

H0
S+
(L/( f1, . . . , fi)L) =

⋃

n≥1

(( f1, . . . , fi)L :L S+
n)/( f1, . . . , fi)L.

Therefore

H0
S+
(L/( f1, . . . , fi)L)a ⊆ (( f1, . . . , fi)L :L S+)/( f1, . . . , fi)L ⊆ H0

S+
(L/( f1, . . . , fi)L).

Hence a((( f1, . . . , fi)L :L S+)/( f1, . . . , fi)L) = a, and the result follows.

The following corollary generalizes [5, Corollary 2.3] to the module case.

Corollary 1. Let g = grade(S+, L). Then:

(i) ai(L) = −∞ for i < g.

(ii) ag(L)≥ −g.

(iii) If H1
S+
(L) 6= 0, then a1(L)≥ −1.



N. Zamani / Eur. J. Pure Appl. Math, 7 (2014), 429-436 432

Proof. We may assume that the base ring S0 is local with infinite residue field. Then from
the graded version of prime avoidance theorem (see for example [2, Proposition 1.5.12]) there
exists an L-sequence f1, . . . , fg of homogeneous elements of S1. Since

( f1, . . . , fg)L :L S+ = ( f1, . . . , fg)L

for i = 1, . . . , g; hence Lemma 2(ii) implies that max{a j(L)+ j : j = 1, . . . , i−1}= −∞. Hence
ai(L) = −∞ for i = 0, . . . , g − 1. As a consequence,

ag(L) + g =max{ai(L) + i : i = 0, . . . , g}= a((( f1, . . . , fg)L :L S+)/( f1, . . . , fg)L)≥ 0.

Therefore, ag(L)≥ −g and (i) and (ii) have been proved.

To prove (iii), set S̄ = S/H0
S+
(S) and L̄ = L/H0

S+
(L). Then it is easy to see that

grade(S̄+, L̄)≥ 1 and H1
S̄+
(L̄)∼= H1

S+
(L) 6= 0. Therefore a1(L) = a1(L̄)≥ −1 by (ii).

Theorem 1. Let f= f1 ∈ S1, . . . , fh ∈ S1 be an S+-filter regular sequence for L. Let b= ( f1, . . . , fh)
be a reduction of S+ with respect to L. Then

reg(L) =max{e(f, L), rb(S+, L)}.

Proof. By Lemma 2 we have

e(f, L) =max{end((( f1, . . . , fi)L :L S+)/( f1, . . . , fi)L) : i = 0, . . . , h− 1}.

Furthermore, rb(S+, L) = end(L/bL) = end((( f1, . . . , fh)L :L S+)/( f1, . . . , fh)L). Therefore

max{e(f, L), rb(S+, L)}=max{end((( f1, . . . , fi)L :L S+)/( f1, . . . , fi)L) : i = 0, . . . , h}
=max{ai(L) + i : i = 0, . . . , h}.

Since reg(L) = max{ai(L) + i : i ≥ 0}, it is enough to show that H i
S+
(L) = 0 for all i > h.

If h = 0, then L is annihilated by some power of S+ and so H i
S+
(L) = 0 for all i > 0. So let

h ≥ 1. By induction, we have H i
S+
(L/ f1 L) = 0 for all i > h− 1. Hence ai(L/ f1 L) = −∞ for

all i > h− 1. By Lemma 1, this implies ai+1(L) = −∞ and H i+1
S+
(L) = 0 for all i > h.

3. Regularity results

In this section, using the ideas of [5], we will show that there is a close relationship be-
tween the invariants ai(Rb(E)) and ai(Gb(E)), from which we can easily derive the formula
reg(R(E)) = reg(G(E)) which is a generalization of that of Ooishi [4] and [5, Theorem 3.1].
For simplicity we shall denote Rb(E) by R(E), Gb(E) by G(E), R(b)+ by R+ and G(b)+ by G+.



N. Zamani / Eur. J. Pure Appl. Math, 7 (2014), 429-436 433

Theorem 2. Let the notation be as in above. Then:

(i) For each i 6= 1, ai(R(E))≤ ai(G(E)).

(ii) ai(R(E)) = ai(G(E)) if ai+1(G(E))≤ ai(G(E)), i 6= 1.

(iii) If H1
G+
(G(E)) 6= 0 or if b ⊆

p

(0 :A E), the statements (i) and (ii) hold for i = 1.

(iv) If H1
G+
(G(E)) = 0 and b 6⊆

p

(0 :A E) then a1(R(E)) = −1.

Proof. We consider the exact sequence

0 −→ R(E)+ −→ R(E) −→ E −→ 0, (1)

where E is considered as a graded R-module concentrated in degree zero.

Since H0
R+
(E)n = 0 for n 6= 0 and H i

R+
(E) = 0 for i ≥ 1, so from the exact sequence (1)

we deduce that H i
R+
(R(E)+)n ∼= H i

R+
(R(E))n for n = 0, i ≥ 2, and for n 6= 0, i ≥ 0. Since

H i
G+
(G(E)) = H i

R+
(G(E)), the exact sequence

0 −→ R(E)+(1) −→ R(E) −→ G(E) −→ 0, (2)

induces the exact sequence

H i
R+
(R(E)+)n+1→ H i

R+
(R(E))n→ H i

R+
(G(E))n→ H i+1

R+
(R(E)+)n+1. (3)

Replacing H i
R+
(R(E)+)n+1 by H i

R+
(R(E))n+1 and setting H i

R+
(G(E)) = 0 whenever that is pos-

sible, we get an epimorphism H i
R+
(R(E))n+1→ H i

R+
(R(E))n for all n≥max{0, ai(G(E)) + 1} if

i = 0,1, and for n ≥ ai(G(E)) + 1 if i ≥ 2. Since H i
R+
(R(E))n = 0 for large values of n, so we

deduce that
H i

R+
(R(E))n = 0

for n ≥ max{0, ai(G(E)) + 1} if i = 0, 1 and for n ≥ ai(G(E)) + 1 if i ≥ 2. From the above
formula immediately we have ai(R(E))≤ ai(G(E)) for i ≥ 2.

For i = 0 we consider two cases. If H0
R+
(G(E)) = 0, then a0(G(E)) = −∞. There-

fore by (4) H0
R+
(R(E))n = 0 for all n ≥ 0. From this it follows that H0

R+
(R(E)) = 0. Hence

a0(R(E)) = −∞ = a0(G(E)). If H0
R+
(G(E)) 6= 0, a0(G(E)) ≥ 0. Hence H0

R+
(R(E))n = 0 for

n≥ a0(G(E)) + 1 by (4), which implies a0(R(E))≤ a0(G(E)). So (i) is proved.

If H1
R+
(G(E)) 6= 0, then a1(G(E)) ≥ −1 by Corollary 1(iii). Hence by (4) H1

R+
(R(E))n = 0

for n ≥ a1(G(E)) which implies a1(R(E)) ≤ a1(G(E)). If b ⊆
p

(0 :A E), then H i
R+
(R(E)) = 0

and H i
R+
(G(E)) = 0 for all i ≥ 1. Hence a1(R(E)) = a1(G(E)) = −∞. So the first part of (iii)

is proved.



N. Zamani / Eur. J. Pure Appl. Math, 7 (2014), 429-436 434

Now we prove (ii) and the second part of (iii). It is sufficient to show that
ai(G(E))≤ ai(R(E)) for i ≥ 0. We may assume that ai(G(E)) 6= −∞. For i = 0, we have either
a1(R(E)) ≤ −1 or a1(R(E)) ≤ a1(G(E)) by (4). For i ≥ 1, we have ai+1(R(E)) ≤ ai+1(G(E))
by (i). Hence the assumption ai+1(G(E)) ≤ ai(G(E)) implies that ai+1(R(E)) ≤ ai(G(E)). Put
n = ai(G(E)). Then H i+1

R+
(R(E)+)n+1

∼= H i+1
R+
(R(E))n+1 = 0. Using this in the exact sequence

(3), we get an epimorphism

H i
R+
(R(E))n −→ H i

R+
(G(E))n.

Since H i
R+
(G(E))n 6= 0, so H i

R+
(R(E))n 6= 0. Therefore, ai(G(E))≤ ai(R(E)).

To prove (iv) we assume that H1
R+
(G(E)) = 0. Then a1(G(E)) = −∞. Hence

a1(R(E)) ≤ −1 by (4). If a1(R(E)) < −1, H1
R+
(R(E))−1 = 0. Since H0

R+
(G(E))−1 = 0, from the

exact sequence (2) we can deduce that H1
R+
(R(E)+)0 = 0. Now, using the exact sequence (1)

we get the exact sequence

H0
R+
(R(E)+)0 −→ H0

R+
(R(E))0 −→ H0

R+
(E) −→ 0.

But since (R(E)+)0 = 0, so H0
R+
(R(E)+)0 = 0. Furthermore, H0

R+
(R(E))0 = H0

b (E) and

H0
R+
(E) = E. Therefore, H0

b (E) = E which is equivalent to the condition bt E = 0 for some

t ≥ 1. Thus if, b 6⊆
p

(0 :A E), we must have a1(R(E)) = −1. Now, the proof of the theorem is
complete.

Corollary 2. Let ` :=max{i : H i
G+
(G(E)) 6= 0}. Then:

(i) a`(R(E)) = a`(G(E)),

(ii) If b ⊆
p

(0 :A E) or `≥ 1, then `=max{i : H i
R+
(R(E)) 6= 0}.

Proof. For i ≥ `, we have ai(G(E)) ≥ ai+1(G(E)) = −∞. Therefore, ai(R(E)) = ai(G(E))
if i 6= 1 by Theorem 2(ii). Hence (i) and(ii) are obvious if ` > 1. It remains to show that
a1(R(E)) = a1(G(E)) if ` = 1 or if ` = 0 and b ⊆

p

(0 :A E). But this follows from Theorem
2(iii).

Corollary 3. With the notation as in above we have

reg(R(E)) = reg(G(E)).

Proof. By Theorem 2(i) we have ai(R(E)) + i ≤ ai(G(E)) + i for i 6= 1. By Theorem 2(iii)
and (iv), either a1(R(E)) + 1≤ a1(G(E)) + 1 or a1(R(E)) + 1= 0≤ reg(G(E)). Therefore,

reg(R(E)) =max{ai(R(E)) + i : i ≥ 0} ≤max{ai(G(E)) + i : i ≥ 0}= reg(G(E)).

To prove reg(G(E))≤ reg(R(E)), let i be maximal such that reg(G(E)) = ai(G(E)) + i. Then
H i

G+
(G(E)) 6= 0 and ai+1(G(E))< ai(G(E)). Now, using Theorem 2(ii),(iii), we get

ai(R(E)) = ai(G(E)). Hence reg(G(E)) = ai(R(E)) + i ≤ reg(R(E)).

In the following we consider R(b) as a subring of the polynomial ring A[t].



REFERENCES 435

Proposition 1. Let f1, . . . , fh be a sequence of elements of b. Then f := f1 t, . . . , fh t is an R(b)+-
filter regular sequence for R(E) if and only if for all large n≥ 1,

[( f1, . . . , fi−1)b
nE :E fi]∩ bnE = ( f1, . . . , fi−1)b

n−1E for i = 1, . . . , h. (4)

If this is the case, then e(f, R(E)) is the least integer r such that (4) holds for all n≥ r + 1.

Proof. The sequence f= f1 t, . . . , fh t is an R(b)+-filter regular sequence for R(E) if and only
if [( f1 t, . . . , fi−1 t)R(E) :R(E) fi t]n is equal to [( f1 t, . . . , fi−1 t)R(E)]n for all large n ≥ 1 and all
i = 1, . . . , h. But the first module is equal to [( f1, . . . , fi−1)bnE :E fi] ∩ bnE and the second is
equal to ( f1, . . . , fi−1)bn−1E. We note that e(f, R(E)) is the least integer r such that the equality
[( f1 t, . . . , fi−1 t)R(E) :R(E) fi t]n = [( f1 t, . . . , fi−1 t)R(E)]n holds for all n≥ r + 1.

Corollary 4. Let a= ( f1, . . . , fh) be a reduction of b with respect to E. Suppose that
f= f1 t, . . . , fh t is an R(b)+-filter regular sequence for R(E). Then

reg(R(E)) =min{r ≥ 0 : r ≥ ra(b, E) and (4) holds for all n≥ r + 1}.

Proof. Let Q = ( f1 t, . . . , fh t). Since a is a reduction of b relative to E, then Q is a reduction
of R(b)+ relative to R(E). Moreover if abnE = bn+1, then QR(b)n+R(E) = R(b)n+1

+ R(E) and
ra(b, E) = rQ(R(b)+, R(E)). By Theorem 1,

reg(R(E)) =max{e(f, R(E)), ra(b, E)}.

Therefore, the result follows from Proposition 1.

Similarly as for Proposition 1, we can prove the following characterization of a homoge-
neous G(b)+ filter regular sequence of degree 1 for G(E). If x ∈ A then x∗ denotes the initial
form of x in G(b).

Proposition 2. Let f1, . . . , fh be elements of b. Then f∗ = f ∗1 , . . . , f ∗h is an G(b)+-filter regular
sequence for G(E)) if and only if for large values of n,

[( f1, . . . , fi−1)b
nE + bn+2E] :E fi ∩ bnE = (( f1, . . . , fi−1)b

n−1E + bn+1E)

for i = 1, . . . , s. If this is the case, e(f∗, G(E)) is the least number r such that the above equality
holds for n≥ r + 1.

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